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3 years ago

5

A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t

he tractor takes 2.5 seconds to react and applies the breaks at a -1.65 m/s/s coming to a stop. How far did the hayride move before coming to a stop?

1 answer:

nexus9112 [7]3 years ago

6 0

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m$

So, the hayride will cover a distance of 68.18 m.

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An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

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Answer:

2.5 m

Explanation:

Weight of billboard worker = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension in rope = 550 N

The sum of the torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$

The position of the person will be 2.5 m

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3 years ago

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

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3 years ago

What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s

Ksivusya [100]

The net force is 270 N

Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 90.0 kg

$a=3.0 m/s^2$

Substituting, we find the net force on the object:

$F=(90.0)(3.0)=270 N$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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3 years ago

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Alex Ar [27]

Answer:

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3 years ago

10points asap <br><br> A force of 30 N acts upon a 7 kg block. Calculate its acceleration.

nekit [7.7K]

Hello! Assuming that the only force acting on the mass is 30N...

Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
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a = 4.2857 m/s^2
a = 4 m/s^2

I hope this helps!

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2 years ago